Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $q = \dfrac{k^2 + 19k + 90}{k + 1} \times \dfrac{-9k - 9}{2k^2 + 20k} $
Answer: First factor the quadratic. $q = \dfrac{(k + 10)(k + 9)}{k + 1} \times \dfrac{-9k - 9}{2k^2 + 20k} $ Then factor out any other terms. $q = \dfrac{(k + 10)(k + 9)}{k + 1} \times \dfrac{-9(k + 1)}{2k(k + 10)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ (k + 10)(k + 9) \times -9(k + 1) } { (k + 1) \times 2k(k + 10) } $ $q = \dfrac{ -9(k + 10)(k + 9)(k + 1)}{ 2k(k + 1)(k + 10)} $ Notice that $(k + 1)$ and $(k + 10)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -9\cancel{(k + 10)}(k + 9)(k + 1)}{ 2k(k + 1)\cancel{(k + 10)}} $ We are dividing by $k + 10$ , so $k + 10 \neq 0$ Therefore, $k \neq -10$ $q = \dfrac{ -9\cancel{(k + 10)}(k + 9)\cancel{(k + 1)}}{ 2k\cancel{(k + 1)}\cancel{(k + 10)}} $ We are dividing by $k + 1$ , so $k + 1 \neq 0$ Therefore, $k \neq -1$ $q = \dfrac{-9(k + 9)}{2k} ; \space k \neq -10 ; \space k \neq -1 $